I have a H11AA1 opto coupler that outputs 5 volts (VCC) with no input and 3 volts with a signal in. I want to use one transistor to change the swing from 5 volts to < 0.5 volts.
I know I can do it with a comparator but is it possible with a single trnsistor?
Thanks
Al
PS Forget my question above. Here are the circuits. The one on the left is used for train detection and I have over a dozen on my layout that work perfectly.
The one on the right is supposed to work the same but is very dependent on the type of DIP bridge rectifier used.
A VM18 works - a B40C800DM and a EDI-PL10 don't work. Seems there is not enough voltage drop
across the two latter bridges to fully turn on the H11AA1.
I agree with Mike. The transistor collector resistor is apparently too small.
According to the data sheet, the maximum "ON" transistor current is about 20% of the LED current you are applying to the coupler. So adjust the collector resistance accordingly.
I'll bump this back up - maybe it'll save some one else some time. I know I spent
a day on it before I decided to change the Bridge rectifier.
BTW the turnon voltage for the H11AA1 is 0.95 volts to 1.1 volts temperature dependent.
I measured 1.2 volts across the VM18 bridge and 0.95 volts across the B40C800DM
which supports my supposition above. Not a very stable design.
Al
The emitters in the opto-coupler are LEDS, and as such, are CURRENT operated devices, not VOLTAGE operated. For normal operation, the CURRENT through the LED should be ~10mA. The forward voltage drop is 1.2 to 1.5V. When you put an appropriate current limiting resistor (one that drops at least 2 to 3V so the current remains more-or-less constant) and add that 2V to the 1.2V to the forward drop across the LED means that you need an minimum voltage of ~3.2V.
What is the role of the "shorted" bridge rectifier? Are you trying to develop enough voltage across the two diode drops to operate the opto-isolator emitter? If so, that is not the way to build a "current sensor".
The emitters in the opto-coupler are LEDS, and as such, are CURRENT operated devices, not VOLTAGE operated. For normal operation, the CURRENT through the LED should be ~10mA. The forward voltage drop is 1.2 to 1.5V. When you put an appropriate current limiting resistor (one that drops at least 2 to 3V so the current remains more-or-less constant) and add that 2V to the 1.2V to the forward drop across the LED means that you need an minimum voltage of ~3.2V.
What is the role of the "shorted" bridge rectifier? Are you trying to develop enough voltage across the two diode drops to operate the opto-isolator emitter? If so, that is not the way to build a "current sensor".
That is exactly the role of the short from + to - . And it works for me in at least 15 circuits. Suggest you see N-Scale Magazine May-June 1993 page 54.
Al
...
That is exactly the role of the short from + to - . And it works for me in at least 15 circuits. Suggest you see N-Scale Magazine May-June 1993 page 54.
Al
I can see using the bridge rectifyer for train detection, But I dont see a use for the one on the right. Just power the opto from the 5 volts using a dropping resistor and forget the bridge rectifyer. Andy
Not complaining now.
This works fine with any of the bridge rectifiers.
Gives 1.8 volts across the bridge/back-to-back diodes
and 40 ma through the H11AA1, two thirds of the max rating.
Fair question. I want to be able to sense with a microcontroller when there is no voltage between the two rails either positive to negative or negative to positive.
It is about the only way to sense externally when a Marklin Z-Scale transfer table has reached a track position.